Lecture 6: Assembly Programs

Transcription

1 Lecture 6: Assembly Programs Today s topics: Procedures Examples Large constants The compilation process A full example 1

2 Procedures Local variables, AR, $fp, $sp Scratchpad and saves/restores, $fp Arguments and returns 2 jal and $ra

3 Procedures Each procedure (function, subroutine) maintains a scratchpad of register values when another procedure is called (the callee), the new procedure takes over the scratchpad values may have to be saved so we can safely return to the caller parameters (arguments) are placed where the callee can see them control is transferred to the callee acquire storage resources for callee execute the procedure place result value where caller can access it return control to caller 3

4 Jump-and-Link A special register (storage not part of the register file) maintains the address of the instruction currently being executed this is the program counter (PC) The procedure call is executed by invoking the jump-and-link (jal) instruction the current PC (actually, PC+4) is saved in the register $ra and we jump to the procedure s address (the PC is accordingly set to this address) jal NewProcedureAddress Since jal may over-write a relevant value in $ra, it must be saved somewhere (in memory?) before invoking the jal instruction How do we return control back to the caller after completing the callee procedure? 4

5 The Stack The register scratchpad for a procedure seems volatile it seems to disappear every time we switch procedures a procedure s values are therefore backed up in memory on a stack Proc A s values High address Proc A Proc B s values Proc C s values Stack grows this way Low address return call Proc B call Proc C return return 5

6 Saves and Restores 6

7 Storage Management on a Call/Return A new procedure must create space for all its variables on the stack Before/after executing the jal, the caller/callee must save relevant values in $s0-$s7, $a0-$a3, $ra, temps into the stack space Arguments are copied into $a0-$a3; the jal is executed After the callee creates stack space, it updates the value of $sp Once the callee finishes, it copies the return value into $v0, frees up stack space, and $sp is incremented On return, the caller/callee brings in stack values, ra, temps into registers The responsibility for copies between stack and registers may fall upon either the caller or the callee 7

8 Example 1 (pg. 98) int leaf_example (int g, int h, int i, int j) { int f ; f = (g + h) (i + j); return f; Notes: In this example, the callee took care of saving the registers it needs. The caller took care of saving its $ra and $a0-$a3. leaf_example: addi $sp, $sp, -12 sw $t1, 8($sp) sw $t0, 4($sp) sw $s0, 0($sp) add $t0, $a0, $a1 add $t1, $a2, $a3 sub $s0, $t0, $t1 add $v0, $s0, $zero lw $s0, 0($sp) lw $t0, 4($sp) lw $t1, 8($sp) addi $sp, $sp, 12 jr $ra Could have avoided using the stack altogether. 8

9 Saving Conventions Caller saved: Temp registers $t0-$t9 (the callee won t bother saving these, so save them if you care), $ra (it s about to get over-written), $a0-$a3 (so you can put in new arguments) Callee saved: $s0-$s7 (these typically contain valuable data) Read the Notes on the class webpage on this topic 9

10 Example 2 (pg. 101) int fact (int n) { if (n < 1) return (1); else return (n * fact(n-1)); Notes: The caller saves $a0 and $ra in its stack space. Temp register $t0 is never saved. fact: slti $t0, $a0, 1 beq $t0, $zero, L1 addi $v0, $zero, 1 jr $ra L1: addi $sp, $sp, -8 sw $ra, 4($sp) sw $a0, 0($sp) addi $a0, $a0, -1 jal fact lw $a0, 0($sp) lw $ra, 4($sp) addi $sp, $sp, 8 mul $v0, $a0, $v0 jr $ra 10

11 Dealing with Characters Instructions are also provided to deal with byte-sized and half-word quantities: lb (load-byte), sb, lh, sh These data types are most useful when dealing with characters, pixel values, etc. C employs ASCII formats to represent characters each character is represented with 8 bits and a string ends in the null character (corresponding to the 8-bit number 0); A is 65, a is 97 11

12 Example 3 (pg. 108) Convert to assembly: void strcpy (char x[], char y[]) { int i; i=0; while ((x[i] = y[i])!= `\0 ) i += 1; Notes: Temp registers not saved. strcpy: addi $sp, $sp, -4 sw $s0, 0($sp) add $s0, $zero, $zero L1: add $t1, $s0, $a1 lb $t2, 0($t1) add $t3, $s0, $a0 sb $t2, 0($t3) beq $t2, $zero, L2 addi $s0, $s0, 1 j L1 L2: lw $s0, 0($sp) addi $sp, $sp, 4 jr $ra 12

13 Large Constants Immediate instructions can only specify 16-bit constants The lui instruction is used to store a 16-bit constant into the upper 16 bits of a register combine this with an OR instruction to specify a 32-bit constant The destination PC-address in a conditional branch is specified as a 16-bit constant, relative to the current PC A jump (j) instruction can specify a 26-bit constant; if more bits are required, the jump-register (jr) instruction is used 13

14 Starting a Program C Program x.c Compiler Assembly language program Assembler x.o Object: machine language module x.s x.a, x.so Object: library routine (machine language) Linker Executable: machine language program a.out Loader Memory 14

15 Role of Assembler Convert pseudo-instructions into actual hardware instructions pseudo-instrs make it easier to program in assembly examples: move, blt, 32-bit immediate operands, etc. Convert assembly instrs into machine instrs a separate object file (x.o) is created for each C file (x.c) compute the actual values for instruction labels maintain info on external references and debugging information 15

16 Role of Linker Stitches different object files into a single executable patch internal and external references determine addresses of data and instruction labels organize code and data modules in memory Some libraries (DLLs) are dynamically linked the executable points to dummy routines these dummy routines call the dynamic linker-loader so they can update the executable to jump to the correct routine 16

17 Full Example Sort in C (pg. 133) void sort (int v[ ], int n) { int i, j; for (i=0; i<n; i+=1) { for (j=i-1; j>=0 && v[j] > v[j+1]; j-=1) { swap (v,j); void swap (int v[ ], int k) { int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; Allocate registers to program variables Produce code for the program body Preserve registers across procedure invocations 17

18 The swap Procedure Register allocation: $a0 and $a1 for the two arguments, $t0 for the temp variable no need for saves and restores as we re not using $s0-$s7 and this is a leaf procedure (won t need to re-use $a0 and $a1) swap: sll $t1, $a1, 2 add $t1, $a0, $t1 lw $t0, 0($t1) lw $t2, 4($t1) sw $t2, 0($t1) sw $t0, 4($t1) jr $ra void swap (int v[], int k) { int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; 18

19 The sort Procedure Register allocation: arguments v and n use $a0 and $a1, i and j use $s0 and $s1; must save $a0 and $a1 before calling the leaf procedure The outer for loop looks like this: (note the use of pseudo-instrs) move $s0, $zero # initialize the loop loopbody1: bge $s0, $a1, exit1 # will eventually use slt and beq body of inner loop addi $s0, $s0, 1 j loopbody1 exit1: for (i=0; i<n; i+=1) { for (j=i-1; j>=0 && v[j] > v[j+1]; j-=1) { swap (v,j); 19

20 The sort Procedure The inner for loop looks like this: addi $s1, $s0, -1 # initialize the loop loopbody2: blt $s1, $zero, exit2 # will eventually use slt and beq sll $t1, $s1, 2 add $t2, $a0, $t1 lw $t3, 0($t2) lw $t4, 4($t2) bgt $t3, $t4, exit2 body of inner loop addi $s1, $s1, -1 exit2: j loopbody2 for (i=0; i<n; i+=1) { for (j=i-1; j>=0 && v[j] > v[j+1]; j-=1) { swap (v,j); 20

21 Saves and Restores Since we repeatedly call swap with $a0 and $a1, we begin sort by copying its arguments into $s2 and $s3 must update the rest of the code in sort to use $s2 and $s3 instead of $a0 and $a1 Must save $ra at the start of sort because it will get over-written when we call swap Must also save $s0-$s3 so we don t overwrite something that belongs to the procedure that called sort 21

22 Saves and Restores sort: addi $sp, $sp, -20 sw $ra, 16($sp) sw $s3, 12($sp) sw $s2, 8($sp) 9 lines of C code 35 lines of assembly sw $s1, 4($sp) sw $s0, 0($sp) move $s2, $a0 move $s3, $a1 move $a0, $s2 # the inner loop body starts here move $a1, $s1 jal swap exit1: lw $s0, 0($sp) addi $sp, $sp, 20 jr $ra 22

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